Page 74 - Modeling of Chemical Kinetics and Reactor Design
P. 74

44    Modeling of Chemical Kinetics and Reactor Design

                                                        Example 1-3
                                The Thermal decomposition of ethane to ethylene, methane, butane,
                              and hydrogen can be expressed by the following mechanism.
                                        
                                 CH →       2 CH * 3                                   (1-190)
                                         k 1
                                     6
                                  2
                                 CH +  C H  →      CH +  C H * 5                       (1-191)
                                               
                                    *
                                               k 2
                                    3
                                                       4
                                                            2
                                           6
                                         2
                                        
                                 CH →       CH  4  +  H *                              (1-192)
                                     *
                                         k 3
                                               2
                                  2
                                     5
                                 H +  C H  →      H +  C H *                           (1-193)
                                              
                                  *
                                              k 4
                                       2  6         2    2  5
                                         k
                                 2CH →       C H 10                                    (1-194)
                                         
                                      *
                                          5
                                   2
                                                4
                                      5
                                Derive a rate law for the formation of ethylene C H  assuming
                                                                                  2
                                                                                    4
                              k <<  k . The free radicals are CH , and C H .
                                                                          *
                                                               *
                                 3   5                         3       2  5
                              Solution
                                                                                             *
                                                                                        *
                                Applying the steady state equations for the free radicals H , CH ,
                                                                                             3
                                      *
                              and C H , the rate of formation of ethylene C H  for a constant
                                                                            2
                                                                               4
                                    2
                                      5
                              volume batch reactor is:
                                           dC
                                               2
                                 + ( r  ) =   CH 4  = kC                                (1-195)
                                   CH 4      dt       3  CH 5 *
                                    2
                                                         2
                              The rate of formation of the free radicals are:
                                ( +r   )  =  dC  CH 3 *  = kC  − k C  C   = 0           (1-196)
                                                                      2
                                                       2
                                   CH 3 *   dt      1  CH 6   2  CH 3 *  CH 6
                                ( +r    )  =  dC CH 5 *
                                               2
                                    2
                                   CH 5 *     dt
                                                                                        (1-197)
                                         = kC    * C    + k C  * C     − k C 2  *  = 0
                                            2
                                                                          5
                                                            4
                                                                   2
                                                    2
                                               CH 3  CH 6     H   CH 6      CH 5
                                                                             2
   69   70   71   72   73   74   75   76   77   78   79